Question: Rewrite the equation by completing the square. $x^{2}-14x+33 = 0$ $(x + $
Solution: Begin by moving the constant term to the right side of the equation. $x^2 - 14x = -33$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-14$, half of it would be $-7$, and squaring it gives us ${49}$. $x^2 - 14x { + 49} = -33 { + 49}$ We can now rewrite the left side of the equation as a squared term. $( x - 7 )^2 = 16$ This is equivalent to $(x+{-7})^2=16$